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Juliette [100K]
3 years ago
15

What is the total number of calories of heat energy absorbed when 10 grams of water is vaporized at its normal boiling point

Chemistry
1 answer:
Darya [45]3 years ago
6 0

<u>Answer:</u> The amount of energy absorbed by water is 5390 Calories

<u>Explanation:</u>

To calculate the amount of heat absorbed at normal boiling point, we use the equation:

q=m\times L_{vap}

where,

q = amount of heat absorbed = ?

m = mass of water = 10 grams

L_{vap} = latent heat of vaporization = 539 Cal/g

Putting values in above equation, we get:

q=10g\times 539Cal/g=5390Cal

Hence, the amount of energy absorbed by water is 5390 Calories

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Explain how you determine the freezing point of a solution that does not have a well-defined transition in the cooling curve.
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This question is asking for a method for the determination of the freezing point in a solution that does not have a noticeable transition in the cooling curve, which is basically based on a linear fit method.

The first step, would be to understand that when the transition is well-defined as the one on the attached file, we can just identify the temperature by just reading the value on the graph, at the time the slope has a pronounced change. For instance, on the attached, the transition occurs after about 43 seconds and the freezing point will be about 4 °C.

However, when we cannot identify a pronounced change in the slope, it will be necessary to use a linear fit method (such as minimum squares) to figure out the equation for each segmented line having a significantly different slope and then equal them so that we can numerically solve for the intercept.

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-3.5 x + 25=-0.52 x + 2\\\\-3.5 x+0.52 x =2-25\\\\x=\frac{-23}{-2.98}=7.72

Next, we plug it in in any of the trendlines to obtain the freezing point as the y-value:

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